"DDoouubbllee ssppeeaakk!!"

Question

Super simple challenge today, or is it?

I feel like we've heard a fair bit about double speak recently, well let's define it in a codable way...

Double speak is when each and every character in a string of text is immediately repeated. For example:

"DDoouubbllee  ssppeeaakk!!"

---

The Rules

• Write code which accepts one argument, a string.
• It will modify this string, duplicating every character.
• Then it will return the double speak version of the string.
• It's code golf, try to achieve this in the smallest number of bytes.
• Please include a link to an online interpreter for your code.
• Input strings will only contain characters in the printable ASCII range. Reference: http://www.asciitable.com/mobile/

---

Leaderboards

Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.

var QUESTION_ID=188988;
var OVERRIDE_USER=53748;
var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;function answersUrl(d){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+d+"&pagesize=100&order=asc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(d,e){return"https://api.stackexchange.com/2.2/answers/"+e.join(";")+"/comments?page="+d+"&pagesize=100&order=asc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(d){answers.push.apply(answers,d.items),answers_hash=[],answer_ids=[],d.items.forEach(function(e){e.comments=[];var f=+e.share_link.match(/\d+/);answer_ids.push(f),answers_hash[f]=e}),d.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(d){d.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),d.has_more?getComments():more_answers?getAnswers():process()}})}getAnswers();var SCORE_REG=function(){var d=String.raw`h\d`,e=String.raw`\-?\d+\.?\d*`,f=String.raw`[^\n<>]*`,g=String.raw`<s>${f}</s>|<strike>${f}</strike>|<del>${f}</del>`,h=String.raw`[^\n\d<>]*`,j=String.raw`<[^\n<>]+>`;return new RegExp(String.raw`<${d}>`+String.raw`\s*([^\n,]*[^\s,]),.*?`+String.raw`(${e})`+String.raw`(?=`+String.raw`${h}`+String.raw`(?:(?:${g}|${j})${h})*`+String.raw`</${d}>`+String.raw`)`)}(),OVERRIDE_REG=/^Override\s*header:\s*/i;function getAuthorName(d){return d.owner.display_name}function process(){var d=[];answers.forEach(function(n){var o=n.body;n.comments.forEach(function(q){OVERRIDE_REG.test(q.body)&&(o="<h1>"+q.body.replace(OVERRIDE_REG,"")+"</h1>")});var p=o.match(SCORE_REG);p&&d.push({user:getAuthorName(n),size:+p[2],language:p[1],link:n.share_link})}),d.sort(function(n,o){var p=n.size,q=o.size;return p-q});var e={},f=1,g=null,h=1;d.forEach(function(n){n.size!=g&&(h=f),g=n.size,++f;var o=jQuery("#answer-template").html();o=o.replace("{{PLACE}}",h+".").replace("{{NAME}}",n.user).replace("{{LANGUAGE}}",n.language).replace("{{SIZE}}",n.size).replace("{{LINK}}",n.link),o=jQuery(o),jQuery("#answers").append(o);var p=n.language;p=jQuery("<i>"+n.language+"</i>").text().toLowerCase(),e[p]=e[p]||{lang:n.language,user:n.user,size:n.size,link:n.link,uniq:p}});var j=[];for(var k in e)e.hasOwnProperty(k)&&j.push(e[k]);j.sort(function(n,o){return n.uniq>o.uniq?1:n.uniq<o.uniq?-1:0});for(var l=0;l<j.length;++l){var m=jQuery("#language-template").html(),k=j[l];m=m.replace("{{LANGUAGE}}",k.lang).replace("{{NAME}}",k.user).replace("{{SIZE}}",k.size).replace("{{LINK}}",k.link),m=jQuery(m),jQuery("#languages").append(m)}}

body{text-align:left!important}#answer-list{padding:10px;float:left}#language-list{padding:10px;float:left}table thead{font-weight:700}table td{padding:5px}

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="https://cdn.sstatic.net/Sites/codegolf/primary.css?v=f52df912b654"> <div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr></tbody> </table> 

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

# Language Name, [Other information] N bytes

where N is the size of your submission. Other information may include flags set and if you've improved your score (usually a struck out number like <s>M</s>). N should be the right-most number in this heading, and everything before the first , is the name of the language you've used. The language name and the word bytes may be links.

For example:

# [><>](http://esolangs.org/wiki/Fish), <s>162</s> 121 [bytes](https://esolangs.org/wiki/Fish#Instructions)

Answer 1

Japt -m, 1 byte

²

Try it online

Answer 2

Japt, 2 bytes

Oliver beat me to the straightforward, 1 byte solution.

íU

Try it

Interleaves the input with itself.

Answer 3

VDM-SL, 28 bytes

f(a)==conc[[x,x]|x in seq a]

Distributed enumeration of the sequence containing sequences of length 2 of repeated elements of a.

Answer 4

Forth (gforth), 40 bytes

: x 0 ?do dup c@ dup emit emit 1+ loop ;

Try it online!

A byte can be saved by removing the ?, but that will break the handling of zero-length input.

Answer 5

CJam, 3 bytes

q:_

Try it online!

Answer 6

><>, 9 bytes

i::0(?;oo

Try it online! I'm reasonably confident that this is optimal (barring solutions which terminate through errors; i:oo would be such a solution.)

Explanation

i::0(?;oo
i             takes one character of input      [c]
 ::           duplicates it twices              [c, c, c]
   0(?        if it's less than 0 (no input)    [c, c]
      ;       terminate
       oo     output twice                      []
              implicitly wrap to the beginning

Answer 7

J, 2 bytes

2#

The verb "#", or "copy", copies its right argument the number of times specified by the left argument where the left argument is greater than or equal to zero and is either a single number or a vector of numbers of the same length as the first axis of the right argument. So, we can copy things "zero" times to remove them:

   1 0 1 0 # 1 2 3 4

returns 1 3.

Try it online: https://tio.run/##y/r//39qcka@gpGyukt@aVJOqkJxQWpitqI6138A

Answer 8

Minkolang v.0.15, 7 bytes

od?.dOO

Try it here!

Explanation

od?.dOO
o            take one character of input
 d           duplicate top of stack
  ?.         terminate if its 0
    dOO      duplicate and output twice

Answer 9

8086/8088 machine code, 9 bytes

I couldn't resist.

ac       lodsb        Read DS:[SI] into AL, then increment SI.
aa       stosb        Write AL into ES:[DI], then increment DI.
3c 00    cmp al, 0    Is AL = 0?
74 03    je +5        If so, jump to the end of this loop.
aa       stosb        Write AL into ES:[DI], then increment DI.
eb f7    jmp -7       Jump to the beginning of this loop.

Assumptions:

• The input string is a null-terminated string of bytes located at DS:SI.
• There is an output string buffer located at ES:DI.
• The output string buffer is large enough to hold the resulting string.
• It's not necessary to tell the calling code the location of the output string buffer. If it is, add 2 bytes: 57 (push di) at the beginning and 5f (pop di) at the end. This assumes that the stack has 2 bytes free.
• This code will be placed inline, so it's unnecessary to use an instruction to return. If this is false, add 1 byte: c3 (ret) at the end (after the 5f, if applicable).

Answer 10

Kotlin, 47 bytes

{(0..it.length-1).fold(""){s,c->s+it[c]+it[c]}}

Try it online!

Answer 11

Bash, 58 39 bytes

Using only shell builtins, no fancy parsers like sed.

58 bytes: Try it online

a="$*";for((i=0;i<${#a};i++)){ x=${a:i:1};echo -n "$x$x";}

47 bytes: (via primo)

for((;i<${#1};)){ x=${1:i++:1};echo -n "$x$x";}

39 bytes: (by manatwork). Try it online

while read -N1 c;do echo -n "$c$c";done

Answer 12

Starry, 14 bytes

` , + + + . .'

Try it online!

Answer 13

33, 8 bytes

P[ktppP]

Try it online! Or not. How do you get a language on TIO?

The input string has to be null-terminated.

Explanation:

P     P  | Get input character
 [     ] | While it isn't a null character:
  kt     | - Put it in the string register
    pp   | - Print it twice

Answer 14

Java (OpenJDK 8), 88 87 bytes

g->{String d ="";for(int i=0;i<g.length();){d+=""+g.charAt(i)+g.charAt(i++);}return d;}

Try it online!

Ungolfed:

g->{
    String d ="";
    for (int i=0;i<g.length();) {
        d+=""g.charAt(i)+g.charAt(i++);
    }
return d;
};

I'm sure there's way to shorten it, at least it was my first attempt in PPCG.

_{-1 for @manatwork}

Answer 15

x86 machine code, 6 bytes

f:
  AC       LODSB
  AA       STOSB
  AA       STOSB
  E2 FB    LOOP  f
  C3       RET

Takes address of input string in SI, length of input string in CX, address of output string in DI. Should work in 16, 32 and 64-bit mode.

Try it online!

Answer 16

Perl 6, 33 bytes

(@*ARGS[0].comb Xx 2).join.say

Comb splits our string into an array. The X meta operator applies the operator to it's right in a cross product fashion. So with the string "Double" it's doing ("D"x2, "o"x2, etc). Then we join the result and say it to print (with a newline).

Answer 17

Bash, 72 bytes

s=$1;e(){ echo -n "${s:$i:1}";};for ((i=0;i<${#s};i++));do e;e;done;echo

If you don't have a bash terminal somewhere, this should work.

Answer 18

C# (Visual C# Interactive Compiler) with Regex^†-flag, 24 bytes

^†: /u:System.Text.RegularExpressions.Regex (flag is not in the header, so it won't mess up the leader-board)

Port of my Java 8 answer, so look there for an explanation.

s=>Replace(s,".","$0$0")

Try it online.

Answer 19

Retina 0.8.2, 6 bytes

.
$&$&

Try it online! Not guaranteed to work with unprintable characters. Previous 8-byte version worked with newlines:

s`.
$&$&

Try it online!

Answer 20

Twig, 66 bytes

This is just a simple "split, loop, output twice".
Splitting by 0 on a string is similar to splitting by '' (empty string).

{%macro a(s)%}{%for c in s|split(0)%}{{c~c}}{%endfor%}{%endmacro%}

You can try it on https://twigfiddle.com/ldxe0c.

Answer 21

Agda, 110 Bytes

Sorry, no online interpreter available because this is more of a proof checker.

So this was quite annoying since I tried not to use the standard library and thus had to explicitly define lists. Chars also don't exist by now, but the function is polymorphic, so it would work in this case.

data L(A : Set): Set where
 N : L A
 C : A → L A → L A
d :{A : _}→ L A → L A
d N = N
d(C x y) = C x(C x(d y))

Answer 22

PHP, 59 bytes

-1 bytes by @manatwork

function($g){foreach(str_split($g)as$v)$e.=$v.$v;return$e;}

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Full program , 44 bytes

foreach(str_split(readline())as$v)echo$v.$v;

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Answer 23

Matlab >= r2015a, 16 bytes

@(x)repelem(x,2)

Credits to Giuseppe

Note: I haven't found an online interpreter for MATLAB; this answer correctly works with MATLAB r2015a.

Matlab <= r2014b, 23 bytes

@(x)char(kron(x,[1,1]))

Note: I haven't found an online interpreter for MATLAB; this answer correctly works with MATLAB r2014b.

Answer 24

Cubix, 9 bytes

oU;?Av/@o

Try it online!

Mapped onto a cube

    o U
    ; ?
A v / @ o . . .
. . . . . . . .
    . .
    . .

Watch it run

• A Put all the input onto the stack
• v Redirect around the cube and start of the print loop
• o? Output the character for the stack value and test value
• @ If test is negative (EOI), halt
• Uo;/ If test is positive, u-turn to the left, output character again, pop from stack and reflect onto the start of the print loop.

Answer 25

Octave, 32 25 bytes

@(x)char(kron(x+0,[1,1]))

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-7 bytes thanks to Giuseppe

Answer 26

Fortran (GFortran), 100 bytes

Another demonstration of fortran's concise and elegant string handling capabilities... not!!

Summary: declare S as a 99 character string. read S in string format '(A)'. iterate over the trimmed string length. print the i'th character a couple of times without a newline (TIL).

Try it Online!

character(99)S;read(*,'(A)')S;do i=1,len_trim(S)
write(*,'(A)',advance="no")S(i:i)//S(i:i);enddo;end

Answer 27

K (ngn/k), 5 bytes

,/2#'

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---

the above is an expression that will take (#) 2 copies of each (') letter and return a list where each element is a 2-char list, e.g.

("DD";"oo";"uu";"bb";"ll";"ee";"  ";"ss";"pp";"ee";"aa";"kk";"!!")

,/ is "join over", and flattens the list. we read the expression as "join over 2 take each string". with the string passed right-side we get:

  ,/2#'"Double speak!"
"DDoouubbllee  ssppeeaakk!!"

Answer 28

AsciiDots, 27 21 bytes

.>#a?**>$_a#
 \---/\/

Note that this halts in an error, which is allowed.

-6 bytes by getting rid of whitespace and not using warps.

Try it online!

Answer 29

Kipple (cipple), 28 bytes

(i>ac<0>b(a-1b+1c+1a?)b>o<c)

It's actually suprisingly difficult to duplicate stack items in Kipple

Expanation

(i>ac<0>b(a-1b+1c+1a?)b>o<c) Full program

(i                         ) Loops until the input stack is empty
 i>a                         Push the top item of the input stack to stack A
    c<a>b                    Push 0's to stack B and C
         (a-1b+1c+1a?)       A loop that basically pushes the top of stack A to stack B and C
                      b>o<c  Push the top of stack B and stack C to the output stack

Try it online!

Answer 30

MathGolf, 2 bytes

I hope this isn't a duplicate. (Great, it isn't.)

_^

Explanation

_  Duplicate the implicit input
 ^ Zip

Try it online!